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External microphone & electrical aspects of design


I am working for a while now on a small project of building an external mic for the iPhone to be hooked in parallel to headphones (via the 3.5mm jack) - basically I want to replace the iPhone original headset microphone with another microphone , but still use the headphones to listen.

link to prototype structure -

some clarifications on the image:

item #4 - iPhone plug 3.5mm

item #3 - a microphone which for now is NOT working since the iPhone seems NOT to identify it (probably impedance issues)

item #2 - simple headphones jack (any headphones with 3.5mm plug can hook there)

I would like to focus my question on the electrical aspects of my project :

known data I gathered so far (please feel free to correct any mistake you identify) -

1. iPhone supply's 1.5v on the TRRS 3.5mm jack.

2. the TRRS plug of iPhone is built of from 4 pins: Left/Right/Ground/Mic


1. What's the power consumption per EACH part of the iPhone headset? Each part means that there are 2 components - headphones & microphone and I need the separated power consumption (especially the microphone!!)

2. I read somewhere that the iPod headphones requires 60mW but I am not sure it's the same , and anyway, we have here the microphone which I assume takes some power, and I need to know how to design it based on iPhone capabilities.

3. Read on an answer previously given to me on this site, that the iPhone identifies external microphone (on the headset for example) only if the impedance is ±1650Ohm - question is : should I plan that the microphone impedance (includes the wire) be 1650Ohm or it means that the whole prototype I have + headphones I will hook to 3.5mm connector (item #2 on the image) should be together 1650Ohm?

4. is this correct to say that the 1.5v supplied by the iPhone 3.5mm audio output, means that the right/left/mic pins are positive contacts with 1.5v vs. the ground pin (means that we have 3 parallel circuits sourced by one 1.5v power source).

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Dizüstü Bilgisayar Bataryaları

An easy fix for a big power boost.

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geva, as a partial answer to your question here is the following. I just finished measuring the output of the audio jack. With a constant sound at full volume I get 1.7Volts AC and 12 milliamps. It will have to be AC since the electrical current flowing through the voice coil changes direction, the coil's polar orientation reverses. this is what makes the cone move and produces sound. therefore your point 4. is not true. The voltage needs to constantly change to produce the different sounds. From the numbers I got, using the equation Power = Current × Voltage P = I × V 12X1.7= 20.4 mW Hope this is a start with your project.

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